Answer:
2.1258 miles from the runway is the airplane at the start of this approach
Step-by-step explanation:
ΔABC
AB = 1,983 feets
[tex]\tan\alpha =\frac{AB}{AC}[/tex]
[tex]\tan10^o=0.1763=\frac{1,983}{AC}[/tex]
[tex]AC=\frac{1,983}{0.1763}=11,247.87 feets[/tex]
1 foot = 0.000189 mile
1,247.87 feets = 1,247.87 × 0.000189 mile = 2.1258 miles
2.1258 miles from the runway is the airplane at the start of this approach
